The radius $r$ of a circle inscribed within three mutually externally tangent circles of radii $a$, $b$ and $c$ is given by
\[\frac{1}{r} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 2 \sqrt{\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc}}.\]What is the value of $r$ when $a = 4$, $b = 9$ and $c = 36$?

[asy]
unitsize(0.15 cm);

pair A, B, C, P;
real r = 1.86308;

A = (0,0);
B = (25,0);
C = intersectionpoint(arc(A,22,0,180),arc(B,27,0,180));
P = intersectionpoint(arc(A,10 + r,0,180),arc(B,15 + r,0,180));

draw(Circle(A,10));
draw(Circle(B,15));
draw(Circle(C,12));
draw(Circle(P,r));
draw(A--(A + (0,-10)));
draw(B--(B + (15,0)));
draw(C--(C + (0,12)));
draw(P--(P + (r,0)));

label("$a$", A + (0,-10)/2, W);
label("$b$", B + (15,0)/2, S);
label("$c$", C + (0,12)/2, W);
label("$r$", P + (r,0), E);
[/asy]
Explanation: We have
\begin{align*}
\frac{1}{r} &= \frac{1}{4} + \frac{1}{9} + \frac{1}{36} + 2 \sqrt{\frac{1}{4 \cdot 9} + \frac{1}{4 \cdot 36} + \frac{1}{9 \cdot 36}} \\
&= \frac{9}{36} + \frac{4}{36} + \frac{1}{36} + 2 \sqrt{\frac{36}{4 \cdot 9 \cdot 36} + \frac{9}{4 \cdot 9 \cdot 36} + \frac{4}{4 \cdot 9 \cdot 36}} \\
&= \frac{14}{36} + 2 \sqrt{\frac{49}{4 \cdot 9 \cdot 36}} \\
&= \frac{7}{9},
\end{align*}so $r = \boxed{\frac{9}{7}}.$